The value of k, If the angle between the straight lines $4 x - y + 7 = 0, k x - 5 y - 9 = 0 i s 45^{0} i s$

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$\frac{- 25}{3}, 3$

$\frac{25}{3}, - 3$

$\frac{- 25}{3}, - 3$

$\frac{25}{3}, 3$

answer is A.

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Detailed Solution

Given lines

$\begin{array}{l} 4 x - y + 7 = 0 \Rightarrow m_{1} = \frac{- 4}{- 1} = 4 \\ a n d k x - 5 y - 9 = 0 \Rightarrow m_{2} = \frac{- k}{- 5} = \frac{k}{5} \\ a n d l e t θ = 45^{0} \\ Tan 45^{0} = |\frac{4 - \frac{k}{5}}{1 + 4 (\frac{k}{5})}| \\ \Rightarrow 1 = |\frac{20 - k}{5 + 4 k}| \end{array}$

$\begin{array}{l} \Rightarrow \frac{20 - k}{5 + 4 k} = \pm 1 \\ ∴ \frac{20 - k}{5 + 4 k} = 1 \frac{20 - k}{5 + 4 k} = - 1 \\ \Rightarrow 20 - k = 5 + 4 k 20 - k = - 5 - 4 k \\ \Rightarrow 20 - 5 = 4 k + k \Rightarrow 20 + 5 = - 4 k + k \\ \Rightarrow 15 = 5 k \Rightarrow - 3 k = 25 \\ \Rightarrow k = 3 \Rightarrow k = \frac{- 25}{3} \end{array}$